Find Sum of a Sequence Defined by Tn = n^2 – (n – 1)^2

A sequence, whose nth term is defined by: T_n = (n)² – (n – 1)².
The task is to calculate the sum of first n term of the sequence i.e find S_n where, S_n = T₁ + T₂ + ….. + T_n-1 + T_n

Examples:

Input : n = 2
Output : 4
T1 = 12 - 02 = 1
T2 = 22 - 12 = 4 - 1 = 3
So, the sum is 4.

Input : n = 1
Output : 1

Method-1:
Simply find each term from T₁, T₂, …, T_n and add them.
This can be done by running a loop i from 1 to n and for each iteration find T_i and add them to a variable.

Implementation in C++:

// CPP Program to find the 
// sum of the first n numbers of the sequence whose 
// nth term is the difference 
// of the squares of n and n - 1
#include <bits/stdc++.h>
using namespace std;

// Return the sum of the sequence
int sumSequence(int n)
{
    int sum = 0;

    // For each ith term, 
    // finding the term and adding it to the result.
    for (int i = 1; i <= n; i++)
    {
        sum += (i * i - (i - 1) * (i - 1));
    }

    return sum;
}

// Driver Program
int main()
{
    int n = 2;

    cout << sumSequence(n) << endl;
    return 0;
}

Output:

4

Method-2:
We are given that: T_n

= (n)^2 - (n - 1)^2 
= n^2 - (n^2 - 2*n + 1) 
= 2*n - 1 

So, nth term can be find by 2*n – 1
Now to calculate S_n

= T1 + T2 + ..... + Tn-1 + Tn 
= 2*1 - 1 + 2*2 - 1 + 2*3 - 1 + ..... 2*n - 1 
= 2*(1 + 2 + .... + n - 1 + n) - n 
= 2*(n*(n + 1)/2) - n 
= n*(n + 1) - n 
= n^2 + n - n 
= n^2  

Implementation in C++:

// CPP Program to find the sum 
// of the first n numbers of the sequence whose
// nth term is the difference 
// of the squares of n and n - 1
#include <iostream>
using namespace std;

// Return the sum of the sequence
int sumSequence(int n)
{
    int sum = 0;

    for (int i = 1; i <= n; i++)
    {
        sum += (i * i - (i - 1) * (i - 1));
    }

    return sum;
}

// Driver Program
int main()
{
    int n = 2;

    cout << sumSequence(n) << endl;
    return 0;
}

Output:

4